Question

in triangle abc ad perpendicular to BC prove that AC square equals to a b square + BC square minus 2 BC multipled by BD

Answer 1

Solution: Mistake in Question , It should be  AC² = AB²+BC² + 2BC×BD

According to the question Triangle is an Obtuse triangle , obtuse angled at B and AD is perpendicular to BC .

To Prove : AC² = AB²+BC² + 2BC×BD

Proof:  In Right angle Triangle ADB we have AB² = AD²+BD² ( by using the Pythagoras Theorem)  --(1)

and in Triangle ADC

AC² = AD²+DC² ( by using the Pythagoras Theorem)

⇒AC² = AD²+(DB+BC)²

⇒AC² = AD²+DB²+BC²+2DB×BC

⇒AC² = AB²+BC²+2BC×BD

NOW from eq. 1 we get AB² = AD²+BD²

hence Proved  AC² = AB²+BC²+2BC×BD

Answer 2

Answer:

$AC^2=AB^2+BC^2-2.BC.BD$

Step-by-step explanation:

Concept:

Pythagors theorem:

In a right angled triangle, square on the hypotenuse is equal to sum of the squares on the other two sides.

In right triangle ADC,

$AC^2=AD^2+CD^2........(1)$

In right triangle ADB,

$AB^2=AD^2+BD^2.........(2)$

From (1)

$AC^2=AD^2+CD^2\\\\AC^2=(AB^2-BD^2)+CD^2\\\\AC^2=(AB^2-BD^2)+(BC-BD)^2\\\\AC^2=AB^2-BD^2+BC^2+BD^2-2.BC.BD\\\\AC^2=AB^2+BC^2-2.BC.BD$